I managed to find
$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-2^{-m}\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}$$
$$=(1-2^{-m})\left[\left(1+\frac m2\right)\zeta(m+1)-\frac12\sum_{i=1}^{m-2}\zeta(i+1)\zeta(m-i)\right]$$
$$+\frac{(-2)^{-m-1}}{(m-1)!}\left[2\gamma\ \psi^{(m-1)}\left(\frac12\right)-\psi^{(m)}\left(\frac12\right)+\lim_{\substack{a\to1/2}}\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$
Where $\overline{H}_k$ is the skew harmonic number, $\gamma$ is Euler–Mascheroni constant, $\zeta$ is the Riemann zeta function and $\psi^{(m)}(a)$ is the Polylogarithm function where
$$\psi^{(m)}\left(\frac12\right)=(-1)^mm!(1-2^{m+1})\zeta(m+1)$$
My question is the closed form above known in the literature? and can we do further simplifications for the limit term to have a cleaner closed form? Also I would like to see different approaches.
Thank you
Proof
$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=1+\sum_{k=2}^\infty\frac{\overline{H}_k}{k^m}=1+\sum_{k=1}^\infty\frac{\overline{H}_{2k}}{(2k)^m}+\sum_{k=1}^\infty\frac{\overline{H}_{2k+1}}{(2k+1)^m}$$
By writing $\overline{H}_{2k}=H_{2k}-H_k$ and $\overline{H}_{2k+1}=H_{2k+1}-H_k$ we have
$$\sum_{k=1}^\infty\frac{\overline{H}_{2k}}{(2k)^m}=\sum_{k=1}^\infty\frac{H_{2n}}{(2n)^m}-\sum_{n=1}^\infty\frac{H_{n}}{(2n)^m}=\frac12\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k^m}+\left(\frac12-2^{-m}\right)\sum_{k=1}^\infty\frac{H_{k}}{k^4}$$
and
$$\sum_{k=1}^\infty\frac{\overline{H}_{2k+1}}{(2k+1)^m}=\color{blue}{\sum_{k=1}^\infty\frac{H_{2k+1}}{(2k+1)^m}}-\sum_{k=1}^\infty\frac{H_k}{(2n+1)^m}$$
$$=\color{blue}{-1+\sum_{n=0}^\infty\frac{H_{2n+1}}{(2n+1)^m}}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}$$
$$=\color{blue}{-1+\frac12\sum_{k=0}^\infty\frac{(-1)^kH_{k+1}}{(k+1)^m}+\frac12\sum_{k=0}^\infty\frac{H_{k+1}}{(k+1)^m}}-\sum_{n=1}^\infty\frac{H_n}{(2n+1)^m}$$
$$=\color{blue}{-1-\frac12\sum_{k=1}^\infty\frac{(-1)^kH_{k}}{k^m}+\frac12\sum_{k=1}^\infty\frac{H_{k}}{k^m}}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}\\$$
Combine the two sums,
$$\Longrightarrow \sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-\sum_{k=1}^\infty\frac{H_k}{(2k+1)^m}\tag1$$
The first sum is well-known
$$\sum_{k=1}^\infty\frac{H_k}{k^m}=\left(1+\frac m2\right)\zeta(m+1)-\frac12\sum_{i=1}^{m-2}\zeta(i+1)\zeta(m-i)$$
For the second sum, from here we have
$$\int_0^1\frac{x^{n}\ln^m(x)\ln(1-x)}{1-x}\ dx=(-1)^{m-1}m!\sum_{k=1}^\infty\frac{H_k}{(k+n+1)^{m+1}}\\=\frac12\frac{\partial^m}{\partial n^m}\left(H_n^2+H_n^{(2)}\right),\quad n\in\mathbb{R}\ge-1,\quad m\in\mathbb{N}$$
Let $m+1\to m$ and $n+1=a$ we get
$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}(H_{a-1}^2+H_{a-1}^{(2)})$$
Substitute $H_{a-1}=\psi(a)+\gamma$ and $H_{a-1}^{(2)}=\zeta(2)-\psi^{(1)}(a)$
$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}((\psi(a)+\gamma)^2+\zeta(2)-\psi^{(1)}(a))$$
Because $m\ge 2$ for convergence, we can ignore the constants $\gamma$ and $\zeta(2)$ on the right side,
$$(-1)^m (m-1)!\sum_{k=1}^\infty\frac{H_k}{(k+a)^m}=\frac12\frac{\partial^{m-1}}{\partial a^{m-1}}(\psi(a)^2-\psi^{(1)}(a)+2\gamma\ \psi(a))$$
$$=\frac12\left[2\gamma\ \psi^{(m-1)}(a)-\psi^{(m)}(a)+\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$
Now take the limit to both sides and let $a\to 1/2$ we get
$$\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}=\frac{(-1)^m}{2(m-1)!}\left[2\gamma\ \psi^{(m-1)}\left(\frac12\right)-\psi^{(m)}\left(\frac12\right)+\lim_{\substack{a\to1/2}}\frac{\partial^{m-1}}{\partial a^{m-1}}\psi(a)^2\right]$$
By combining the results of the two sums, the closed form follows.
Note
I am tagging " integration" as logarithmic integrals and harmonic series are strongly related.
Best Answer
Yes, a closed form in the literature is known. For $m \geqslant 2$ it is: $$\sum_{k = 1}^\infty \frac{\overline H_k}{k^m} = \zeta (m) \log 2 - \frac{1}{2} m \zeta (m + 1) + \eta (m + 1) + \frac{1}{2} \sum_{i = 1}^m \eta (i) \eta (m - i + 1).$$ Here $\eta (s) = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^s} = (1 - 2^{1 - s}) \zeta (s)$ is the Dirichlet eta function and $\zeta (s) = \sum_{n = 1}^\infty \frac{1}{n^s}$ is the Riemann zeta function.
References:
R. Sitaramachandrarao, "A formula of S. Ramanujan," Journal of Number Theory, 25, 1-19 (1987). See Theorem 3.5 on page 9.
Philippe Flajolet and Bruno Salvy, "Euler sums and contour integral representations," Experimental Mathematics, 7(1), 15-35 (1998). See Theorem 7.1 (i) on page 32.